10.21 Vector Spaces

Definition and several examples

\(\R^2\) is a vector space

​

Definition of vector space

Let \(\mathbb{F}\) be a field. A vector space over \(\mathbb{F}\) or a \(\mathbb{F}\)-vector space is a set \(V\) with a sum, \(+:V\times V\rightarrow V\) and a multiplication by scalars, \(\mathbb{F}\times V\rightarrow V\) satisfying the following:

1. The sum is associative and commutative

2. There is a neutral element for that, denoted 0

3. For every \(v\in V,\exists w\in V\) such that \(v+w=0\)​

4. \(1\cdot v=v,\forall v\in V\)​

5. \(\lambda\cdot(v+w)=\lambda\cdot v+\lambda\cdot w,\forall\lambda\in\mathbb{F},\forall v,w\in V\) field to vector

6. \(\left(\lambda+\mu\right)\cdot v=\lambda\cdot v+\mu\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V\) field to vector

7. \(\lambda\cdot\left(\mu\cdot v\right)=\left(\lambda\cdot\mu\right)\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V\) field to product in F

Example

1. \(V=\R^2\), \((a,b)+(c,d)=(a+c,b+d)\), \(\lambda(a+b)=(\lambda\cdot a,\lambda\cdot b)\) addition and product in real numbers

2. \(V=\mathbb{F}^2\), \((a,b)+(c,d)=(a+c,b+d)\), \(\lambda(a+b)=(\lambda\cdot a,\lambda\cdot b)\) .....in the field

3. \(V=\R^n=\{(a_1,a_2,a_3,...,a_n):a_i\in \R\}\)​

4. \(V=\mathbb{F}^n=\{(a_1,a_2,a_3,...,a_n):a_i\in \mathbb{F}\}\)

Remark

There is one neutral element!

Each \(v\in V\) has one opposite, \(-v\)

Notation

\(v-w=v+(-w)\)

Example

\(\mathbb{F}^\infty=\{(a_1,a_2,...,a_n,...):a_i\in \mathbb{F}\}=\{f:\N\rightarrow \mathbb{F}\}\)

\((a_1,a_2,...,a_n,...)+(b_1,b_2,...,b_n,...)=(a_1+b_2,a_2+b_2,...,a_n+b_n,...)\)​

\(\lambda\cdot(a_1,a_2,...,a_{n},...)=(\lambda a_1,\lambda a_2,...,\lambda a_{n},...)\)​

Some arithmetic properties

1. \(0\cdot v=0\) (is a vector), \(\forall v\in V\)

   Proof

   \(0\cdot v=(0+0)\cdot v=0\cdot v+0\cdot v\)​

   Then \(0\cdot v+(-0\cdot v)=0\cdot v+0\cdot v+(-0\cdot v)\)​

   Then \(0=0\cdot v\)​

2. \((-1)\cdot v=-v,\forall v\in V\)​

   Proof

   \((-1)\cdot v+v=(-1)\cdot v+1\cdot v=v\cdot\left(1+\left(-1\right)\right)=v\cdot0=0\)​

3. \(-(-v)=v,\forall v\in V\)​

   Since \(v+(-v)=0\), then \(-v\) is the additive inverse of \(v\)​

   Thus \(v=-(-v)\)​

Example

1. \(V=\R^3\), \(u=(1,0,-1),v=(2,1,1),w=(-1,1,0)\), then compute \(2u-v+w=2\cdot(1,0,-1)+(-2,-1,-1)+(-1,1,0)=(-1,0,-3)\in\mathbb{R}^3\)

   ‍

   \(\exists \lambda_1,\lambda_2\) such that \(\lambda_1u+\lambda_2v=\left(3,\frac12,-\frac32\right)\)?

   Since the second column of \(\mu\) is zero, then \(\lambda_2=\frac12\)

   Then consider the first column \(\lambda_1+\frac12\cdot2=3\Rightarrow\lambda_1=2\)

   Then check the third column, yes!

   ‍

   Are there \(\lambda,\mu\) such that \(\lambda\cdot u+\mu\cdot v=w\) ? no

   There is three dimension, but the left hand only can ensure a plane, \(w\) is possible not on the plane.

2. \(W=\Z_5^2,\mu=(3,2),v=(2,2)\)​

   Is it possible to write every \(v\in V\) as \(\lambda\cdot u+\mu\cdot v\)?

   Given \((a,b)\in\mathbb{Z}_5^2,\lambda\left(3,2\right)+\mu\left(2,2\right)=\left(a,b\right)\), then \(\left(3\lambda+2\mu,2\lambda+2\mu\right)=\left(a,b\right)\)

   \(\begin{cases}3\lambda+2\mu=a\\ 2\lambda+2\mu=b\end{cases}\), then \(\begin{cases}\lambda=a-b\\ \mu=\frac{3b}{2}-a=4b+4a\end{cases}\)​

3. \(V=\mathbb{C}^2,u=\left(1+i,1-i\right),v=\left(2i,2\right)\)​

   Are there \(\lambda,\mu\) such that \(\lambda\cdot u+\mu\cdot v=w\) ? no

   Observe that \((1+i)(1+i,1-i)=(2i,2)=v\), \(u\) and \(v\) are in the same line.

   ​

   collinear

Remark

\(V=\mathbb{F}^n,v_i=(0,...,1(\text{第i位,其余为0}),...0),i=1,2,...,n\)

\((a_1,a_2,...,a_n)=a_1(1,0,....0)+a_2(0,1,0,....0)+...+a_n(0,0,....1)\)​

Further examples

\(V=\mathbb{F}^{S}=\{f:S\rightarrow\mathbb{F}\}\) (\(S\) is a set)

\((f+g)(s)=f(s)+g(s),\forall s\in S\)

\((\lambda\cdot f)(s)=\lambda f(s),\forall s\in S\)​

Remark

\(S=\{1,2\},\mathbb{F}^{S}\sim\mathbb{F}^2\)

\(S=\{1,2,\ldots,n\},\mathbb{F}^{S}\sim\mathbb{F}^{n}\)

\(S=\N,\mathbb{F}^{S}\sim\mathbb{F}^{\infty}\)

Polynomial and matrices

\(V=\mathbb{F}[x]=\{\text{polynomials with coefficients in }\mathbb{F}\}\) with \(+\) and \(\times\) by scalars you know

\(V=M_{n\times m}\left(\mathbb{F}\right)\) with \(+\) and \(\times\) by scalars we defined

Question

Are there \(p_1(x),...,p_m(x)\) such that any \(q(x)\) is \(q(x)=a_1p_1\left(x\right)+\cdots+a_{m}p_{m}\left(x\right)\)?

No, polynomial is unbounded, we can always find a bigger one

Remark

\(\mathbb{F}=\mathbb{F}[x]_0\subseteq\mathbb{F}\left\lbrack x\right\rbrack_1\subseteq\mathbb{F}\left\lbrack x\right\rbrack_2\subseteq\ldots\subseteq\mathbb{F}\left\lbrack x\right\rbrack_{m}\subseteq\ldots\)

\(\mathbb{F}[x]=\bigcup_{m=0}^{\infty}\mathbb{F}[x]_m\)

\(\mathbb{F}\left\lbrack x\right\rbrack_{m}=\{\text{polynomial of degree at most m}\}\) on \(\mathbb{F}\)-vector space

Linear combination

Let \(V\) be on \(\mathbb{F}\)-vector space. Given a list of vectors \(v_1,v_2,...,v_m\), a linear combination of them is a vector of the form \(w=\lambda_1\cdot v_1+\lambda_2\cdot v_2+\cdots+\lambda_{m}\cdot v_{m}\) for some \(\lambda_1,\lambda_2,\ldots,\lambda_{m}\in\mathbb{F}\)​

‍